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andre [41]
3 years ago
5

A model of a statue is

Mathematics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

32

Step-by-step explanation:

<u>10  divided  by  5/2  or  2  and  1/2  is  4   4  x  8  =  32.</u>

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Joe owns a stock which has probability .5 of going up. This morning, he bought a ticket in a lottery game which gives him a prob
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Answer:

The probability that Joe's stock will go up and he will win in the lottery  is 0.00005.

Step-by-step explanation:

Let the events be denoted as:

<em>X</em> = the stock goes up

<em>Y</em> = Joe wins the lottery

Given:

P (X) = 0.50

P (Y) = 0.0001

The events of the stock going up is not dependent on the the event of Joe winning the lottery.

So the events <em>X</em> and <em>Y</em> are independent of each other.

Independent events are those events that can occur together at the same time.

The joint probability of two independent events <em>A</em> and <em>B </em>is,

P(A\cap B)=P(A)\times P(B)

Compute the value of P (<em>X ∩ Y</em>) as follows:

P(X\cap Y)=P(X)\times P(Y)=0.50\times 0.0001=0.00005

Thus, the probability that Joe's stock will go up and he will win in the lottery  is 0.00005.

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3 years ago
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At a college, 69% of the courses have final exams and 42% of courses require research papers. Suppose that 29% of courses have a
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Answer:

a) 0.82

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Step-by-step explanation:

We are given that

P(F)=0.69

P(R)=0.42

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a)

P(course has a final exam or a research paper)=P(F or R)=?

P(F or R)=P(F)+P(R)- P(F and R)

P(F or R)=0.69+0.42-0.29

P(F or R)=1.11-0.29

P(F or R)=0.82.

Thus, the the probability that a course has a final exam or a research paper is 0.82.

b)

P( NEITHER of two requirements)=P(F' and R')=?

According to De Morgan's law

P(A' and B')=[P(A or B)]'

P(A' and B')=1-P(A or B)

P(A' and B')=1-0.82

P(A' and B')=0.18

Thus, the probability that a course has NEITHER of these two requirements is 0.18.

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3 years ago
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Compare and contrast the expression 2+x and 2+3
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